Question

Find the length and breadth of the rectangle whose area is 6x^2-29x+30. Also find its perimeter.​

Answer 1

Let's assume the length of the rectangle is l and the breadth is b. So, the area of the rectangle is given by l * b = 6x^2 - 29x + 30.

Expanding the area expression, we get l * b = 6x^2 - 29x + 30
=> lb = 6x^2 - 29x + 30
=> lb - 30 = 6x^2 - 29x
=> lb - 30 + 29x = 6x^2
=> (l + b)x - (l - b) = 6x^2

Comparing the coefficients on both sides, we get:

l + b = 6x
l - b = -29

Adding the above two equations, we get 2l = 6x - 29
=> l = (3x - 14.5)

Substituting the value of l in the second equation, we get

b = (-29 + 3x - 14.5)
=> b = -29.5 + 3x

Since the length and breadth of the rectangle are given by expressions, we cannot determine their exact numerical values.

The perimeter of the rectangle is given by 2 * (l + b) = 2 * (3x - 14.5 + (-29.5 + 3x)) = 2 * (-11 + 6x) = -22 + 12x

Answer 2

Answer: l = (2x-3) ; b = (2x-10) ; Perimeter = (10x - 26)

Explanation:

l * b = area

area = 6x^2 - 29x + 30

l * b = 6x^2 - 29x + 30

Split the middle term of 6x^2 - 29x + 30

6x^2 - 20x - 9x + 30

2x(3x-10) -3(3x-10)

(2x-3) (3x-10)

l = (2x-3)

b = (3x-10)

Perimeter of a Rectangle = 2(l+b)

= 2 ((2x-3) + (3x-10))

= 2( 2x-3 + 3x-10 )

= 2 (5x -13)

= (10x - 26)cm